厦门市2016-2017学年度第一学期高二年级质量检测数学(理科)试题参考答案及评分标准.pdf

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1、厦门市2016-2017学年度第一学期高二年级质量检测数学（理科）试题参考答案及评分标准一、选择题：本大题共12小题，每小题5分，共60分．1~5：ADCDB6~10：CBCAC11~12：BD第12题解析：连结PF，1∵OFOFOPPFc，OFP60，1222∴FPF90，PF3c，122c2又∵PF2ac，∴2ac3c，31，2a3122bc2∴11(31)233，22aa2222x0y0xy设P(x,y)，M(x,y)，N(x,y)为椭圆C上的点，则1，1，00002222aba

2、b222b222b22即y(ax)，y(ax)，0202aa222b22b22b2222(ax)(ax0)(x0x)2yyyyyya2a2a2b000kK(233)NPNM2222222xxxxxxxxxxa00000∵k3，∴K23.NPNM2b（另解：取PN中点Q，kKkK，转化为中点弦问题，使用点差法即可）NPNMNPOQ2a二、填空题：本大题共4小题，每小题5分，共20分．22313．若mn0，则mn014．12715．516．2第16题解析：S2SBD2

3、CDABDACDAD是BAC的角平分线，由角平分线定理，得AB2AC在RtAHD中，AD2AH，则ADH30设CDx，ACy法一：由余弦定理：2221(2y)(2x)在ABD中，cos1504y2221yx33在ACD中，cos30解方程组：xBC2y62法二：设由BAD，由正弦定理：2y1在ABD中，，sin150sin(30)y1ACD中sin30sin(150)22解得：cos33sin，结合sincos1，7321解得sin，cos141

4、4高二数学（理科）试题参考答案及评分标准第1页共7页21y，6yx由正弦定理，ACD中，sin30sin33解得xBC623法三：在RtADH中，DH，由勾股定理：22321223212在RtAHC中，y(x)()，在RtAHB中，(2y)(2x)()222233解方程组：xBC62三、解答题：本大题共6小题，共70分．17．本小题考查正、余弦定理、三角形面积公式、两角和三角公式；考查计算求解能力、推理论证能力能力；考查方程思想.本小题满分10分．abc解：法一：（Ⅰ）由正弦定理：2R，sinAsi

5、nBsinC有a2RsinA，b2RsinB，c2RsinC则2cosA(2RsinAcosB2RsinBcosA)2RsinC2cosA(sinAcosBsinBcosA)sinC·····························································1分2cosAsin(AB)sinC···············································································2分sin(AB)s

6、in(C)sinC2cosAsinCsinC·························································································3分0C有sinC02cosA1解得A··········································4分3（Ⅱ）由（Ⅰ）知A，又a7322由余弦定理得：bcbc49（*）········································6分1ABC的面

7、积SbcsinA103，即bc40（**）········································7分2222由（*）（**）得，bcbc(bc)3bc49，··············································8分解得bc13，································································································9分ABC的周长为abc20．····

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